Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
GMAT Club

 It is currently 22 Mar 2017, 21:28

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A bag of 10 marbles contains 3 red marbles and 7 blue

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 37547
Followers: 7390

Kudos [?]: 99199 [0], given: 11008

Re: 3 red marbles and 7 blue marbles [#permalink]

### Show Tags

07 Mar 2012, 17:10
rohitgoel15 wrote:
A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Case 1: Only 2 marbles are blue. Prob: 7/10 * 6/9 * 3/8
Case 2: All 3 marbles are blue. Prob: 7/10 * 6/9 * 5/8

The point is that BBR case can occur in 3 different ways: BBR, BRB, and RBB. So you should multiply 7/10 * 6/9 * 3/8 by 3.

This is discussed here: a-bag-of-10-marbles-contains-3-red-marbles-and-7-blue-56728.html#p677059 and here: a-bag-of-10-marbles-contains-3-red-marbles-and-7-blue-56728.html#p677083
_________________
Senior Manager
Joined: 07 Nov 2009
Posts: 312
Followers: 9

Kudos [?]: 561 [0], given: 20

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

08 Mar 2012, 00:02
1
This post was
BOOKMARKED
Owsum Bunuel ... Thanks !
Senior Manager
Joined: 19 Oct 2010
Posts: 271
Location: India
GMAT 1: 560 Q36 V31
GPA: 3
Followers: 7

Kudos [?]: 77 [0], given: 27

Re: Combinatorics - at least, none .... [#permalink]

### Show Tags

01 Apr 2013, 09:51
Bunuel wrote:
Bullet wrote:
Thanks Bunuel. I posted the solution from the thread as i was confused with the solution.

So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole?

Thanks

To make it simple: suppose we have the jar of 10 marbles - 5 red, 2 blue and 3 green. If five marbles are selected at random, what is the probability that two will be red, one blue and two green?

We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is $$\frac{5!}{2!2!}$$.

Hence the answer for the above question would be $$\frac{5!}{2!2!}*\frac{5}{10}*\frac{4}{9}*\frac{2}{8}*\frac{3}{7}*\frac{2}{6}$$.

If the question were: three marbles are selected at random, what is the probability that all three will be red?

RRR can occur only in one way: RRR, so the probability would be $$\frac{5}{10}*\frac{4}{9}*\frac{3}{8}$$.

You can check the Probability and Combination chapters in the Math Book (link below) for more.
Also check my posts at:
probability-colored-balls-55253.html#p637525
4-red-chips-and-2-blue-chips-85987.html#p644603
probability-qs-attention-88945.html#p671958
p-c-88431.html?highlight=probability+of+occurring+event
probability-88069.html?highlight=probability+of+occurring+event
combination-problem-princenten-review-2009-bin-4-q2-87673.html?highlight=probability+of+occurring+event

That was an awesome explanation. Thank you!
_________________

petrifiedbutstanding

Current Student
Joined: 06 Sep 2013
Posts: 2031
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 65

Kudos [?]: 621 [0], given: 355

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

08 Feb 2014, 14:21
bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?

In number 4 we need to specify that probability of EXACTLY two marbles are blue. Otherwise it would be the same as 3 no?

Just wondering
Cheers
J
Math Expert
Joined: 02 Sep 2009
Posts: 37547
Followers: 7390

Kudos [?]: 99199 [0], given: 11008

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

09 Feb 2014, 01:32
Expert's post
1
This post was
BOOKMARKED
jlgdr wrote:
bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?

In number 4 we need to specify that probability of EXACTLY two marbles are blue. Otherwise it would be the same as 3 no?

Just wondering
Cheers
J

No. If it meant at least two, then it would be written that way.
_________________
Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 58 [0], given: 23

Re: Combinatorics - at least, none .... [#permalink]

### Show Tags

20 Apr 2014, 15:05
Bunuel wrote:
We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is $$\frac{5!}{2!2!}$$.

RRR can occur only in one way: RRR, so the probability would be \frac{5}{10}*\frac{4}{9}*\frac{3}{8}.

Can someone please talk about this a little? I understand that the marbles can be arranged in various ways RBRGG etc. but how do we come up with the $$\frac{5!}{2!2!}$$ equation?

EDIT: To add to the question above, why can RRR only occur in one way, isn't that 3 separate ways as well? Meaning, R_1, R_2, R3 -- there are 3 different R's and therefore 3 different ways?
Manager
Joined: 12 May 2013
Posts: 84
Followers: 2

Kudos [?]: 37 [0], given: 12

Re: Combinatorics - at least, none .... [#permalink]

### Show Tags

21 Apr 2014, 01:48
Bunuel wrote:
Bullet wrote:
Can any body please explain Question No.3 using probability

Why we need to add twice
7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15

Thanks and appreciated

The solution you are posting for the third question is not right. Below is the solution of this question using the "probability".

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

Probability of at least two marble are blue is the sum of the two probabilities:

A. Two marbles are blue and one is red - BBR. This can occur in $$\frac{3!}{2!}=3$$ # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. $$3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}$$;

B. All three marbles are blue - BBB. This can occur only one way, namely BBB. $$\frac{7}{10}*\frac{6}{9}*\frac{5}{8}$$

So $$P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}$$

Bunuel, i just have 1 silly question, here why cant we do: 3!/2!*0.7^2*0.3 , i mean it is not specified in the question whether we replace the ball after selecting it
i get confused in such situations, a lot !
Math Expert
Joined: 02 Sep 2009
Posts: 37547
Followers: 7390

Kudos [?]: 99199 [0], given: 11008

Re: Combinatorics - at least, none .... [#permalink]

### Show Tags

21 Apr 2014, 02:05
Bunuel wrote:
Bullet wrote:
Can any body please explain Question No.3 using probability

Why we need to add twice
7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15

Thanks and appreciated

The solution you are posting for the third question is not right. Below is the solution of this question using the "probability".

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

Probability of at least two marble are blue is the sum of the two probabilities:

A. Two marbles are blue and one is red - BBR. This can occur in $$\frac{3!}{2!}=3$$ # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. $$3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}$$;

B. All three marbles are blue - BBB. This can occur only one way, namely BBB. $$\frac{7}{10}*\frac{6}{9}*\frac{5}{8}$$

So $$P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}$$

Bunuel, i just have 1 silly question, here why cant we do: 3!/2!*0.7^2*0.3 , i mean it is not specified in the question whether we replace the ball after selecting it
i get confused in such situations, a lot !

If it were with replacement it would be specified. Or let me put it this way: proper GMAT question would make it clear whether it's with or without replacement case.
_________________
Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 58 [0], given: 23

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

21 Apr 2014, 18:21
Thanks Bunuel, very clear!

That leads me to my second question -- sorry it's a little long winded.

When we solve this using probability, let's take question 3 for example(10 Marbles, 3R, 7B, if 3 are selected what is the prob that at least 2 will be blue). Following that, if I use probability, it is clear that the solution is P(BBR)+P(BRB)+P(RBB)+P(BBB).

On the other hand, if we use combinatorics, according to the solutions posted above, why aren't permutations taken into account. What I mean by that is, the correct solution is (7C2 x 3C1 + 7C3)/10C3 which implies that it's BBR OR BBB. Why are we ignoring BRB and RBB -- shouldn't it be ((7C2 x 3C1)3) + 7C3)/10C3?
Current Student
Status: The Final Countdown
Joined: 07 Mar 2013
Posts: 297
Concentration: Technology, General Management
GMAT 1: 710 Q47 V41
GPA: 3.84
WE: Information Technology (Computer Software)
Followers: 5

Kudos [?]: 72 [0], given: 444

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

30 Nov 2014, 00:43
In a similar question.

In a blue jar there are red, white and green balls. The probability of drawing a red ball is 1/5. The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10. What is the probability of drawing a white ball?

a) 1/5.
b) ½.
c) 1/3.
d) 3/10.
e) ¼.

the OA is 1/2..which is just 1/10=1/5*x,so x=1/2..
my question is that returning the ball does not figure into the question in anyway?
Math Expert
Joined: 02 Sep 2009
Posts: 37547
Followers: 7390

Kudos [?]: 99199 [0], given: 11008

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

30 Nov 2014, 05:33
Ralphcuisak wrote:
In a similar question.

In a blue jar there are red, white and green balls. The probability of drawing a red ball is 1/5. The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10. What is the probability of drawing a white ball?

a) 1/5.
b) ½.
c) 1/3.
d) 3/10.
e) ¼.

the OA is 1/2..which is just 1/10=1/5*x,so x=1/2..
my question is that returning the ball does not figure into the question in anyway?

The probability of drawing a red ball is 1/5: P = (red)/(total) = 1/5;
The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10: (red)/(total)*(white)/(total) = 1/10.

Since (red)/(total) = 1/5, then 1/5*(white)/(total) = 1/10 --> (white)/(total) = 1/2.

Hope it's clear.
_________________
Intern
Joined: 20 Dec 2014
Posts: 22
Followers: 0

Kudos [?]: 2 [0], given: 31

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

12 Mar 2015, 09:04
nevergiveup wrote:
bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

1- [3!/(2!1!)]/[10!/(2!8!)]=14/15

2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?

[3!/(2!1!)]/[10!/(2!8!)]=1/15

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15

Personally, I really don't know what is the difference between question 3 & 4.

Having trouble with this. For the first question the way that I am attempting to reason through it is 10!/2!8! . 2! represents the 2 marbles selected and 8! is for the other 8 NOT selected. I then have probability first one is not blue ,3/10, times this and answer subtracted from 1. So 1- [10!/2!/8!(3/10)]
Verbal Forum Moderator
Joined: 02 Aug 2009
Posts: 4382
Followers: 371

Kudos [?]: 3817 [0], given: 106

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

12 Mar 2015, 09:12
GMAT01 wrote:
nevergiveup wrote:
bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

1- [3!/(2!1!)]/[10!/(2!8!)]=14/15

2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?

[3!/(2!1!)]/[10!/(2!8!)]=1/15

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15

Personally, I really don't know what is the difference between question 3 & 4.

Having trouble with this. For the first question the way that I am attempting to reason through it is 10!/2!8! . 2! represents the 2 marbles selected and 8! is for the other 8 NOT selected. I then have probability first one is not blue ,3/10, times this and answer subtracted from 1. So 1- [10!/2!/8!(3/10)]

hi GMAT01,
you do not require the coloured portion above...
wherever the questions of this form are there , the best is to take probability none is there and it seems you begun that way but went wrong mid way..
prob of both not being blue means both are red..
prob of first red=3/10 and second red=2/9.....
prob of both red 3/10*2/9=6/90..
so prob of atleast one blue=1-6/90=84/90...
hope its clear
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Intern
Joined: 20 Dec 2014
Posts: 22
Followers: 0

Kudos [?]: 2 [0], given: 31

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

12 Mar 2015, 10:34
Chetan2u thank you for clarifying - I get it now. I was just confusing myself trying to approach this through combinatorics. I just made the problem more complex for me to understand instead of sticking to a simple method.
Intern
Joined: 10 May 2013
Posts: 7
GMAT 1: Q43 V28
Followers: 0

Kudos [?]: 0 [0], given: 170

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

23 Sep 2015, 07:37
Bunuel wrote:
fxsunny wrote:
powerka wrote:
3) Pick 3. P of picking at least 2 blue?
Combinatorics: P(2b&1r)+P(3b) -> (7C2 x 3C1 + 7C3)/10C3
Answer: 49/60 (7/15 and 42/90 are wrong; do the math)

#3. P(at least 2 blues) = P(exactly 2 blues) + P (3 blues)

P (exactly 2 blues)= $$7C2/10C3$$=$$21/120$$
P (3 blues) = $$7C3/10C3$$=$$35/120$$

Therefore,
P(at least 2 blues) = 21/120 + 35/120 = 56/120 = 7/15

I see that the commented section above has an extra x 3C1 that's probably the cause for my discrepancy. Why do I need to multiply $$7C2$$ by 3C1 to get P (exactly 2 blues)? Or, is my answer $$7/15$$correct?

That's because you are picking 3 marbles and if you pick 2 blue then the third one must be red: BBR. Ways to pick one red marble out of 3 is $$C^1_3$$.

Complete solution using combinations: $$P(R\geq{2})=P(R=2)+P(R=3)=\frac{C^2_7*C^1_3}{C^3_{10}}+\frac{C^3_7}{C^3_{10}}$$.

Hope it's clear.

Hello, sir. Could you you please expain why we should assume that 7 blue marbles are not identical?
As I understand, when we deal with ditinct objects, say 7 distinct people A, B, C, D, E, F, and want to choose 2 of them, we find it by 7C2= 7!/2!5!.
But how can be 7 blue marbles be distinct? shouldn't their 2-place combination equal to 1?
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 14364
Followers: 601

Kudos [?]: 174 [0], given: 0

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

24 Oct 2016, 02:59
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 25 Jan 2013
Posts: 14
Concentration: General Management, Entrepreneurship
Followers: 1

Kudos [?]: 3 [0], given: 663

A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

14 Nov 2016, 10:50
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
For above question, can you show me how to solve by
1- p(all blue)
as I am getting different answer by doing 1- (7/10*6/9) = 8/15
Intern
Joined: 12 Dec 2016
Posts: 12
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: A bag of 10 marbles contains 3 red marbles and 7 blue [#permalink]

### Show Tags

21 Jan 2017, 20:42
how's about with replacement, how can i solve a similar problem given that the marble is put back?
Re: A bag of 10 marbles contains 3 red marbles and 7 blue   [#permalink] 21 Jan 2017, 20:42

Go to page   Previous    1   2   [ 38 posts ]

Similar topics Replies Last post
Similar
Topics:
4 A bag contains blue, red and green marbles only. 3 29 Oct 2016, 09:39
33 Bag A contains red, white and blue marbles such that the red 15 08 Sep 2010, 20:56
1 A bag contains 6 red marbles,9 blue marbles,and 5 green marbles.You wi 2 25 Aug 2010, 19:22
22 Bag A contains red, white and blue marbles such that the red 30 29 Jun 2007, 15:38
10 Bag A contains red, white and blue marbles such that the red to white 9 01 Jul 2008, 11:25
Display posts from previous: Sort by